Transcript Activity 34: - Mathematical sciences

ACTIVITY 34
Review (Sections 3.6+3.7+4.1+4.2+4.3)
Problems 3 and 5:
Sketch the graph of the function by transforming an appropriate
function of the form y = xn. Indicate all x- and y-intercepts on
each graph.
P( x)  ( x  2)
3
y - intercept
y xx3 23
y
y  x  2
3
x0
P0  (0  2)3  8
x - intercept
y0
0  ( x  2)3
0  ( x  2)3
0  ( x  2)
 2,0
x  2
0,8
P( x)  2 x  8
4
No x – intercepts
y – intercepts is (0,8)
y = x4
y = 2x4
y = 2x4 + 8
Problems 11, 13, and 15:
Match the polynomial function with the below graphs
P( x)  x( x2  4) R( x)   x5  5x3  4 x
T ( x)  x 4  2 x3
T ( x)  x 4  2 x3
R( x)   x5  5x3  4 x
P( x)  x( x2  4)
Problems 21:
Sketch the graph of the polynomial function. Make sure your
graph shows all intercepts and exhibits the proper end behavior.
P( x)  x  3x  23x  2
x - intercepts/ zero's
x3
The leading term is 3x3
x  2
2
x
3
y - intercept
x0
P(0)  0  30  230  2
  32 2
 12
Problem 31:
Factor the polynomial P(x) = −x3 + x2 + 12x and use the
factored form to find the zeros. Then sketch the graph.
P( x)   x3  x 2  12x
  x x 2  x 12
  xx  4x  3


Zeros
x0
x0
 x  4  0
x4
x  3  0
x  3
Problem 3:
Let P(x) = x4 − x3 + 4x + 2 and Q(x) = x2 + 3.
(a) Divide P(x) by Q(x).
(b) Express P(x) in the form P(x) = D(x) · Q(x) + R(x).
x2 – x – 3
x2 +
x4 − x3 + 4x + 2
3
Q (x )
–xx4 –+ 3x
3x22
P( x )  Q ( x ) D( x )  R( x )
x4  x3  4 x  2  x2  x  3x2  3  7 x  11
− x3 – 3x2 + 4x + 2
+– x33
+
– 3x
3x
– 3x2 + 7x + 2
– 3x22
+3x
+
–9
7x + 11
R (x )
Problem 9:
Find the quotient and remainder using long division for the expression
x3  6 x  3
x2  2 x  2
P( x )  Q ( x ) D( x )  R( x )
x +2
x2 – 2x+2
P( x ) Q ( x ) D( x )  R( x )

D( x )
D( x )

x3 + 6x + 3
–xx3 +
– 2x
2x22 +– 2x
Q ( x ) D( x ) R( x )
R( x )

 Q( x) 
D( x )
D( x )
D( x )
x3  6 x  3
8x  1
 x  2  2
2
x  2x  2
x  2x  2
Q (x )
2x2 + 4x + 3
– 2x2 –+ 4x
4x +– 4
8x – 1
R (x )
Problem 21:
Find the quotient and remainder using synthetic division for the
expression
3 1
x3  8x  2
x3
1
Q( x)  x  3x  1
R( x )  1
2
0
8
2
3
9
3
3
1
1
Q (x )
R (x )
Problem 35:
Use synthetic division and the Remainder Theorem to evaluate
P(c) if P(x) = 5x4 + 30x3 − 40x2 + 36x + 14 and c = −7.
7
5
5
30  40
14
 35 35
35  497
5 5
71  483
Q (x )
P( c )  R( c )
P( 7)   483
36
R (x )
Problem 43:
Use the Factor Theorem to show that x − 1 is a factor of
P(x) = x3 − 3x2 + 3x − 1.
1
1 3
1
3 1
1
2
1
2
1
0
Showing that x = 1 is a zero.
Therefore, (x – 1) is a factor
Problem 53:
Find a polynomial of degree 3 that has zeros 1, −2, and 3,
and in which the coefficient of x2 is 3.
a x  1 x  2 x  3
ax  2 x  x  2x  3
2
ax  x  2x  3
ax3  3x2  x2  3x  2 x  6
2
ax3  2 x 2  5x  6
ax  2ax  5ax  6a
3
2
 2a  3
3
a
2
3
x  1x  2x  3
2
Problem 3:
List all possible rational zeros given by the Rational Zeros Theorem (but don’t
check to see which actually are zeros).
R( x)  2 x5  3x3  4 x2  8
divisors of 8   1,  2,  4,  8
p

 1,  2
divisors of 2
q
possible rational zeros:
 1,  1 ,  2,  4,  8
2
Problems 13 and 23:
Find all rational zeros of the polynomial.
P( x)  x3  3x  2
divisors of 2   1, 2
p

1
divisors of 1
q
possible rational zeros:
x
3
 1,  2
 3  2 1 1
 3x  2  x  x  2 x  (1)
x12  x 12 x  12
1
1
1
02
1
2 4
1
0
1
3 2
1
1  2
2
0


x3  3x  2  x2  x  2 x  1
x
2
 x  2  x  2x  1
x3  3x  2  x  2x  1x  1
Consequently, the zero’s
are x = 2 and x = – 1
P( x)  x 4  6 x3  7 x2  6 x  8
divisors of 8   1,  2,  4,  8
p

divisors of 1
q
1
possible rational zeros:  1,
 2,  4,  8
P( x)  x 4  6 x3  7 x2  6 x  8
1
1
1
possible rational zeros:  1,
6
7
6
8
1
7
14
8
7
14
8
0


3
2
x 4  6 x 3  7 x 2  6 x  8  x  7 x  14x  8 x  1
So we need only factor
x
3
 7 x2  14x  8
 2,  4,  8
possible rational zeros:  1,  2,  4,  8
possible rational zeros:  1,
 2,  4,  8
x
3
 7 x2  14x  8
1 1
1
possible rational zeros:  1,
 2,  4, 8
7
14 8
1  6  8
6
8
0


 x  6 x  8x  (1)x  1
 x  6 x  8x  1x  1
x  6x  8 x  4x  1
3
2
x 4  6 x 3  7 x 2  6 x  8  x  7 x  14x  8 x  1
2
2
So we need only factor
2
x4  6x3  7 x2  6x  8  x  4x  2x  1x  1
Consequently, the roots are x = -4, x = -2, x = -1, and x = 1